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Mathematics 248
Topics in Combinatorial Mathematics
1st Semester AY 2006-2007
1. Find the number of ways to choose a pair {a, b} of distinct numbers from the set {1, 2, . . . , 50} such that (i) |a − b| = 5 (ii) |a − b| ≤ 5 2. There are 12 students in a party. Five of them are girls. In how many ways can these 12 students be arranged in a row if (i) there are no restrictions? JS The number of ways to arrange the 12 students is 12! = P12 12 . (ii) the 5 girls must be together (forming a block)? JS The number of ways to arrange five girls within a block is 5!, while the number of ways to arrange 8 objects — 7 boys and 1 block of girls — is 8!, thus by M.P., the number of ways that the students can be arranged is 5! · 8!. (iii) no 2 girls are adjacent? 12−5+1 JS Given that H12 is the number of ways that five places can be chosen amongst 5 = 5 twelve, so that no two of the places are consecutive, which will be the positions for the girls, 5! ways to arrange the five girls amongst those positions and 7! ways to arrange the seven boys amongst the remaining positions, by M.P., the number of ways that the students can be arranged is 5! · 7! · 85 . (iv) between two particular boys A and B, there are no boys but exactly 3 girls? JS There are 2! ways to arrange A and B, 53 ways to choose the three girls amongst the five that would be between A and B, 3! to arrange these three chosen girls, and 8! ways to arrange the block of five students (A, B and the three girls between them) with the remaining seven students. By M.P., the number of ways that the students can be arranged is 2!· 53 ·3!·8!. 3. m boys and n girls are to be arranged in a row, where m, n ∈ N. Find the number of ways this can be done in each of the following cases: (i) There are no restrictions; JS The number of ways to arrange m + n people in a row is (m + n)! = Pm+n m+n . (ii) No boys are adjacent (m ≤ n + 1); JS There are Hm+n = m+n−m+1 ways of choosing m places among m + n such that no two m m of these places are consecutive, which will be the positions of the boys, m! ways to arrange the boys among the chosen positions and n! ways to arrange the girls among the remaining positions, by M.P., the number of ways to arrange the students is m! · n! · n+1 m . (iii) The n girls form a single block; JS There are n! ways to arrange the girls within their block and (m + 1)! ways to arrange m + 1 objects — the m boys and the block of girls — by M.P., the number of ways to arrange the students is n! · (m + 1)!. (iv) A particular boy and a particular girl must be adjacent. JS There are 2! ways to arrange the boy and the girl within their block, and (m + n − 1)! ways to arrange m + n − 2 + 1 objects — the boy-and-girl block and the m + n − 2 other students — by M.P., the number of ways to arrange the students is 2! · (m + n − 1)!. 4. How many 5-letter words can be formed using A, B, C, D, E, F, G, H, I, J, (i) if the letters in each word must be distinct? JS The number of arrangements of five distinct letters from among the ten letters is P10 5 = 10! = 10 · 9 · 8 · 7 · 6. 5! (ii) if, in addition, A, B, C, D, E, F can only occur as the first, third of fifth letters while the rest as the second or fourth letters? JS For the odd-numbered letters, the choices for the letters are 6, 5 and 4, respectively; for the even-numbered letters, the choices for the letters are 4 and 3, repectively. By M.P., the number of five-letter words that can be constructed is 6 · 4 · 5 · 3 · 4. Exercises 1
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Mathematics 248
Topics in Combinatorial Mathematics
1st Semester AY 2006-2007
5. Find the number of ways of arranging the 26 letters in the English alphabet in a row such that there are exactly 5 letters between x and y. 6. Find the number of odd integers between 3000 and 8000 in which no digit is repeated. JS There are two types of numbers that satisfy the conditions: Type 1: Numbers with odd thousands digits: the thousands digit must be 3, 5 or 7. This will leave only 4 possible digits for the ones digit, and 8 · 7 choices for the middle two digits. There are 3 · 8 · 7 · 4 = 672 such numbers. Type 2: Numbers with even thousands digits: the thousands digit must be 4 or 6. This will leave 5 possible digits for the ones digit, and 8 · 7 choices for the middle two digits. There are 2 · 8 · 7 · 5 = 560 such numbers. Thus, there are 672 + 560 = 1232 such numbers satisfying the conditions. 7. Evaluate 1 · 1! + 2 · 2! + 3 · 3! + · · · + n · n! where n ∈ N. MS Let k = 1 · 1! + 2 · 2! + 3 · 3! + · · · + n · n!. Note that 1+k = 1! + 1 · 1! +2·2!+3·3!+· · ·+n·n! = 2! + 2 · 2! +3·3!+· · ·+n·n! = · · · = (n+1)·n! = (n+1)!. | {z } | {z } 2·1!=2!
3·2!=3!
Thus k = 1 · 1! + 2 · 2! + 3 · 3! + · · · + n · n! = (n + 1)! − 1. JS
1 · 1! + 2 · 2! + 3 · 3! + · · · + n · n! 1! + 2! + 3! + · · · + n! 2! + 3! + 4! + · · · + (n + 1)! − 1! − 2! − 3! − · · · − n! − 1! + (n + 1)!
+
8. Evaluate
2 n 1 + + ··· + (1 + 1)! (2 + 1)! (n + 1)!
where n ∈ N. MS Let k =
1 2 n + + ··· + . Note that (1 + 1)! (2 + 1)! (n + 1)!
+
Thus k +
2 1 + + ··· + (1 + 1)! (2 + 1)! 1 1 + + ··· + (1 + 1)! (2 + 1)! 2 3 + + ··· + (1 + 1)! (2 + 1)!
n (n + 1)! 1 (n + 1)! n+1 (n + 1)!
=
1 1 1 + + ··· + 1! 2! n!
1 1 1 1 1 1 1 2 n 1 + +· · ·+ = + + · · · + and k = + + · · · + = 1− . 2! 3! (n + 1)! 1! 2! n! 2! 3! (n + 1)! (n + 1)!
JS Prove, by induction,
1 2 k−1 1 + + ··· + =1− : (1 + 1)! (1 + 1)! k! k!
1 1 1 = = 1 − . 2! 2 2! induction: Assume that the formula holds for k = n, then for k = 2:
1 2 n−1 n + + ··· + + (1 + 1)! (1 + 1)! n! (n + 1)! | {z }
Exercises 1
1 n (n + 1)! − (n + 1) + n + = n! (n + 1)! (n + 1)!
=
1−
=
(n + 1)! − 1 1 =1− . (n + 1)! (n + 1)! Page 2 of 32
Mathematics 248
Topics in Combinatorial Mathematics
1st Semester AY 2006-2007
9. Prove that for each n ∈ N, (n + 1)(n + 2) · · · (2n) n
is divisible by 2 . 10. Find the number of common positive divisors of 1040 and 2030 . JS Since 1040 = 240 · 540 and 2030 = 260 · 530 , their common positive divisors are of the form 2a · 5b , where 0 ≤ a ≤ 40 and 0 ≤ b ≤ 30. Thus there is a bijection between the common positive divisors of the numbers and the coordinate pairs (a, b) satisfying the given conditions; both sets have 41 · 31 = 1271 elements. 11. In each of the following, find the number of positive divisors of n (inclusive of n) which are multiples of 3: (i) n = 210; JS n = 3 · 7 · 5 · 2, thus the each positive divisor of n which is amultiple of 3 is the product of 3 and a subset of the set {2, 5, 7}; thus, there are 30 + 31 + 32 + 33 = 8 positive divisors of 210 that are multiples of 3. (ii) n = 630; JS n = 3 · 7 · 5 · 3 · 2, thus the each positive divisor of n which is a multiple of 3is the product of 3 and a subset of the set {2, 3, 5, 7}; thus, there are 40 + 41 + 42 + 43 + 44 = 16 positive divisors of 630 that are multiples of 3. (iii) n = 1512000. JS n = 3 · 7 · 52 · 32 · 25 , thus the each positive divisor of n which is a multiple of 3 is the product of 3 and a multi-subset of the multi-set {5 · 2, 2 · 3, 2 · 5, 7}; thus, there are 6 · 3 · 3 · 2 = 108 positive divisors of 1512000 that are multiples of 3. 12. Show that for any n ∈ N, the number of positive divisors of n2 is always odd. k2 kr 2 2k1 2k2 2kr k1 KM Let n ∈ N with prime factorization Qr n = p1 p2 · · · pr . Then, n = p1 p2 · · · pr . The 2 number of positive divisors of n is i=1 (2ki +1) which is odd since (2ki +1) is odd and the product of odd numbers is odd.
13. Show that the number of positive divisors of “111 . . . 1}” is even. | {z 1992
TT, JA Using some basic divisibility rules, we know that 111 . . 1111} is divisible by 3 but not 9. | . {z 1992
Thus, we can express 111 . . 1111} as | . {z 1992
111 . . 1111} = 3r. | . {z 1992
Let the prime factorization of r be r = pa1 1 pa2 2 · · · pakk
where
a1 , a2 , . . . , ak , k ∈ N, p1 , p2 , . . . , pk are prime numbers
Since 3r is not divisible by 9, r will not be divisible by 3. Hence, none of p1 , p2 , . . . , pk will actually be a 3. The prime factorization of 111 . . 1111} is simply given by | . {z 1992
111 . . 1111} = 31 pa1 1 pa2 2 · · · pakk . | . {z 1992
And thus the number of positive divisors that it has will be (1 + 1)(a1 + 1)(a2 + 1) · · · (ak + 1). Exercises 1
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Mathematics 248
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1st Semester AY 2006-2007
We need not concern ourselves with the actual prime factors and their corresponding exponents, because as seen in the expression above, the product is always even. Hence, the number of positive divisors for 111 . . 1111} is always even. | . {z 1992
14. Let n, r ∈ N with r ≤ n. Prove each of the following identities: (i) Pnr = nPn−1 r−1 , n! (n − 1)! =n = nPn−1 r−1 . (n − r)! [(n − 1) − (r − 1)]! (ii) Pnr = (n − r + 1)Pnr−1 , n − r + 1 n! n! n! = = (n − r + 1) = (n − r + 1)Pnr−1 . JS Pnr = (n − r)! n − r + 1 (n − r)! [n − (r − 1)]! n (iii) Pnr = Pn−1 , where r < n, n−r r n (n − 1)! n n! = = Pn−1 . JS Pnr = (n − r)! n − r [(n − 1) − r]! n−r r (iv) Prn+1 = Pnr + rPnr−1 , JS JS Pnr =
Pnr + rPnr−1
= = =
n! n! n! · (n − r + 1) + r · n! +r = (n − r)! [n − (r − 1)]! (n − r + 1)! n! · n − n! · r + n! + r · n! n! · (n + 1) n! · (n + 1) (n + 1)! = = = (n − r + 1)! [(n + 1) − r]! [(n + 1) − r]! [(n + 1) − r]! Pn+1 . r
is the number of arrangements of r distinct objects taken from n + 1 objects. This JS Pn+1 r number of arrangements may also be counted by considering two cases: Case 1: The (n + 1)th object is not among the r distinct objects arranged. Therefore, the r objects must be from the n other objects, which has Pnr arrangements. Case 2: The (n + 1)th object is among the r distinct objects arranged. The (n + 1)th object can be placed in any of the r positions in the arrangement, and the remaining r − 1 distinct objects are taken from the n remaining objects. There are rPnr−1 such arrangements. By A.P., the sum Pnr + rPnr−1 is also the number of arrangements of r distinct objects taken = Pnr + rPnr−1 . from n + 1 objects, and therefore Pn+1 r r (v) Prn+1 = r! + r(Pnr−1 + Pn−1 r−1 + · · · + Pr−1 ). JS Using (iv),
− Pnr =⇒ Pn+1 Pn+1 = Pnr + rPnr−1 r r n−1 n n−1 n Pr = Pr + rPr−1 =⇒ Pr − Pn−1 r n−1 n−1 Pr = Pn−2 + rPn−2 − Pn−2 r r r−1 =⇒ Pr .. . Pn+1 r
= r! +
Pr+1 = Prr + rPrr−1 =⇒ r r + Pn−1 + · · · + P ) ⇐= r−1 r−1
r(Pnr−1
= rPnr−1 = rPn−1 r−1 = rPn−2 r−1 .. .
Pr+1 − Prr = rPrr−1 r n+1 r Pr − Prr = rPnr−1 + rPn−1 r−1 + · · · + rPr−1
15. In a group of 15 students, 5 of them are female. If exactly 3 female students are to be students are to be selected, in how many ways can 9 students be chosen from the group (i) to form a committee? JA, JS From the given, we have 5 female students, thus, we have 10 male students, i.e., 15 − 5 = 10. Now, we have the following restrictions: that are exactly 3 female students are to be selected. So, 5 . (15a) 3 Exercises 1
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Mathematics 248
Topics in Combinatorial Mathematics
1st Semester AY 2006-2007
Since we need only exactly 3 students from the group of five female students and knowing that a committee is composed of 9 students, we have to choose 6 male students from the remaining 10 male students. For that case we have, 10 . (15b) 6 Hence, by M.P., multiplying (15a) and (15b) above, there are 5 10 = 2, 100 3 6 ways of choosing 9 students from to group of 15 to form a committee. (ii) to take up 9 different posts in a committee? JA, JS To answer the second question: There are 9! ways for each students to take up different posts in a committee since every students in the committee could possibly take up these 9 position. Thus, by M.P., we have 5 10 9!. 3 6 16. Ten chairs have been arranged in a row. Seven students are to be seated in seven of them so that no two students share a common chair. Find the number of ways this can be done if no two empty chairs are adjacent? JS There are 7! ways that the seven students can be arranged in a selection of seven chairs. At most one empty chair can be placed between any two chairs that seat students, and there may be an empty chair at the ends of the row, so there are eight possible places to have the three empty chairs, thus 83 ways to select the positions of the empty chairs in relation to the chairs that seat students. By M.P., the number of ways that the students can be seated is 7! · 83 . 17. Eight boxes are arranged in a row. In how many ways can five distinct balls be put into the boxes if each box can hold at most one ball and no two boxes without balls are adjacent? JS There are H83 = 8−3+1 ways to select three empty boxes among the eight boxes so that no 3 two empty boxes are next to each other, and 5! ways to arrange the balls among the non-empty boxes. By M.P., the number of ways to place the balls in the boxes are 5! · 63 . 18. A group of 20 students, including 3 particular girls and 4 particular boys, are to be lined up in two rows with 10 students each. In how many ways can this be done if the 3 particular girls must be in the front row while the 4 particular boys be in the back? JS There are 10! ways to arrange the students in the front row, 10! ways to arrange the students in the back row, and 13 6 ways to choose from the 20 − 3 − 4 students (other than the particular boys and girls) six other students to accompanythe four boys in the back row. By M.P., the number of ways to line up the students is 10! · 10! · 13 6 . 19. In how many ways can 7 boys and 2 girls be lined up in a row such that the girls must be separated by exactly 3 boys? JS There are 2! to arrange the two girls, 73 ways of choosing the three boys who will be between the girls from among the seven boys, 3! ways to arrange the chosen boys between the girls, and (4 + 1)! ways to arrange the four remaining boys and the block of five people that have the girls at the ends. By M.P., the number of ways to line up the boys and girls is 2! · 73 · 3! · 5!. 20. In a group of 15 students, 3 of them are female. If at least one female student is to be selected, in how many ways can 7 students be chosen from the group (i) to form a committee? LL, FO There are 31 12 there are 32 12 5 ways to form the committee with exactly 1 female; 5 ways to form the committee with only two females; and there are 33 12 4 ways to form the committee with the three females. By AP, the total number of ways to form the committee with Exercises 1
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Mathematics 248
least one female is
Topics in Combinatorial Mathematics
3 12 + 32 12 1 5 6 15 7 ways to form
+
3 3
12 4
1st Semester AY 2006-2007
= 5, 643.
LL, FO There are the committee of 7 students with no restrictions and 12 7 ways to form the committeewith no females. Thus the number of ways to form the committee 12 with at least 1 female is 15 7 − 7 . (ii) to take up 7 different posts in a committee? LL, FO Since there are 7! ways for each student to take up different posts in a committee and student in the committee could possibly take the 7 positions. Thus, by MP, we have every 12 7! 15 − = (5040)(5643) = 28, 440, 720. 7 7 21. Find the number of (m + n)-digit binary sequences with m 0s and n 1s such that no two 1s are adjacent, where n ≤ m + 1. 22. Two sets of parallel lines with p and q lines each are shown in the following diagram:
p | {z }
q
Find the number of parallelograms formed by the lines. MS Since a parallelogram is formed by two intersecting pairs of parallel lines, choosing two lines amongst the p horizontal lines and two lines amongst the q diagonal lines creates a parallelogram, and each parallelogram in the figure is created by a pair amongst the p horizontal lines and a pair amongst the q diagonal lines, the number of parallelograms in the diagram is equal to the number of ways that a pair of horizontal lines and a pair of diagonal lines can be selected, which is p2 2q . 23. There are 10 girls and 15 boys in a junior class, and 4 girls and 10 boys in a senior class. A committee of 7 members is to be formed from these 2 classes. Find the number of ways this can be done if the committee must have exactly 4 senior students and exactly 5 boys. FO Restriction: 1) 7 members of the committee form 2 classes, 2) exactly 4 senior students, 3) exactly 5 boys. Consider the ff. cases: Case 1: Since we only need 7 members of the committee from 2 classes having exactly 4 senior students and exactly 5 boys, we choose 4 boys from the senior class thus we write 10 4 . Because the committee must have exactly 5 boys, we have to pick 7 boys from the junior class. We cannot pick another boy from the senior class since we are restricted that exactly 4 senior students must be in the committee and it must have exactly 5 boys. Thus we choose 1 boy from the junior class having 15 boys and write 15 to choose 2 female 1 . Now, we need students from the junior class to complete the committee, we have 10 . 2 15 10 By (MP), 10 = 141, 750 ways of choosing 4 senior boys, 1 boy from the junior class 4 1 2 and 2 girls from the junior class. Case 2: Choosing 3 boys from the senior class 1 girl from the senior class 2 boys from the junior class 1 girl from the junior class 4 15 10 By (MP), 10 3 1 2 1 = 504, 000 ways.
Exercises 1
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Mathematics 248
Topics in Combinatorial Mathematics
1st Semester AY 2006-2007
Case 3: Choosing 2 boys from the senior class 2 girl from the senior class 3 boys from the junior class 4 15 By (MP), 10 2 2 3 = 122, 850 ways. Therefore, by (AP), there are 141, 750 + 504, 000 + 122, 850 = 768, 600 ways a committee of 7 members formed from 2 classes having exactly 4 senior students and exactly 5 boys. 24. A box contains 7 identical white balls and 5 identical black balls. They are to be drawn randomly, one at a time without replacement, until the box is empty. Find the probability that the 6th ball drawn is white, while before that exactly 3 black balls are drawn. PB
5 2
6 12 5
2 =
25 . 132
Note that the first factor in the numerator is the number of ways of arranging the first five balls, three of which are black and two are white. The second factor in the numerator is the number of ways of arranging the last six balls four of which are black and two are white. Also, the sixth ball is fixed to be a white ball. The denominator is just the total number of arranging twelve balls, seven of which are black and five are white. 25. In each of the following cases, find the number of shortest routes from O to P in the street network shown below: P A O
r
r rB
rC
r
(i) The routes must pass through the junction A; PB We first count the number of ways to go from point O to point A, then from point A to 5 8 point P . This is given by: = 560. 2 3 (ii) The routes must pass through the street AB; PB We first count the number of ways to go from point O to point A, then from point B to 5 7 point P . This is given by: = 350. 2 3 (iii) The routes must pass through junctions A and C; PB We first count the number of ways to go from point O topoint A, then from point A to 5 4 4 point C, then finally from point C to point P . This is given by: = 240. 2 1 2 (iv) The street AB is closed. PB We first count the number of ways to go from point O to point P, then we subtract the 13 5 7 number of ways we pass through AB. This is given by: − = 937. 5 2 3 26. Find the number of ways of forming a group of 2k people from n couples, where k, n ∈ N with 2k ≤ n, in each of the following cases: (i) There are k couples in such a group; (ii) No couples are included in such a group; (iii) At least one couple is included in such a group; (iv) Exactly two couples are included in such a group.
Exercises 1
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Mathematics 248
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1st Semester AY 2006-2007
27. Let S = {1, 2, . . . , n + 1} where n ≥ 2, and let T = {(x, y, z) ∈ S 3 | x < z and y < z}. Show by counting |T | in two two different ways that n X n+1 n+1 2 k = |T | = +2 . 2 3 k=1
KM Method 1: If z ≥ 2, then there are z − 1 choices for both x and y. Thus, |T | =
Pn+1 i=2
(i − 1)2 =
Pn
k=1
k2 .
Method 2: Consider 2 disjoint cases. Case 1: x = y. Then, there are n+1 such triples. Choose 2 numbers from S, the larger one is z, 2 and the smaller one is x and y. Case 2: x 6= y. Then, there are 2 n+1 such triples. First, choose 3 numbers from S, the largest 3 one is z. We multiply by 2 because there are two possible positions for the remaining two numbers from S (after having set z). Thus, |T | = n+1 + 2 n+1 2 3 . 28. Consider the following set of points on the xy-plane: A = {(a, b) | a, b ∈ Z, 0 ≤ a ≤ 9 and 0 ≤ b ≤ 5}. Find PB Note that we have 10 a coordinates, and 6 b coordinates. (i) the number of rectangles whose vertices are points in A; PB To form a rectangle, we simply select two points from the 10 a coordinates, and two from 10 6 the 6 b coordinates. This is given by: = 675. 2 2 (ii) the number of squares whose vertices are points in A. PB We count the number of squares using cases: 1 × 1 square: 2 × 2 square: 3 × 3 square: 4 × 4 square: 5 × 5 square:
5 · 9 such squares 4 · 8 such squares 3 · 7 such squares 2 · 6 such squares 1 · 5 such squares
Hence, there are 115 squares. 29. Fifteen points P1 , P2 , . . . , P15 are drawn in the plane in such a way that besides P1 , P2 , P3 , P4 , P5 which are drawn collinear, no other 3 points are collinear. Find (i) the number of straight lines which pass through at least 2 of the fifteen points; PB We basically count the number of ways of connecting any two points. Now, since five points are collinear, then we subtract the number of ways of connecting any two of these five points, and add 1 for the line formed by these five collinear points. This is given by: 15 5 − + 1 = 96. 2 2 (ii) the number of triangles whose vertices are 3 of the 15 points. PB We basically count the number of ways of connecting any three points to form a triangle. Now, since five points are collinear, then we subtract the number of ways of connecting any three of these five points since those three will not form a triangle. This is given by: 15 5 − = 445. 3 3 Exercises 1
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30. In each of the following 6-digit natural numbers: 333333, 225522, 118818, 707099, every digit in the number appears at least twice. Find the number of such 6-digit natural numbers. 31. In each of the following 7-digit natural numbers: 1001011, 5550000, 3838383, 7777777, every digit in the number appears at least 3 times. Find the number of such 7-digit natural numbers. MS A 7-digit must begin with a nonzero digit, giving nine possible options. In relation to the first digit, there are three cases to consider: Case 1: The first digit is repeated twice in the next six digits. The four remaining digits will be the same, so the possibilities of this case are determined by the location of the two repetitions of the first digit, and the choice of the remaining digit. Hence, this case has 9 · 62 · 9 elements. Case 2: The first digit is repeated thrice in the next six digits. The three remaining digits will be the same, so the possibilities of this case are determined by the location of the three repetitions of the first digit, and the choice of the remaining digit. Hence, this case has 9 · 63 · 9 elements. Case 3: The first digit is repeated in the next six digits. This case has 9 elements. So there are a total of 9 · 62 · 9 + 9 · 63 · 9 + 9 = 2844 such 7-digit natural numbers. 32. Let X = {1, 2, 3, . . . , 1000}. Find the number of 2-element subsets {a, b} of X such that the product a · b is divisible by 5. PB Any product of two numbers is divisible by 5 if at least one of the two numbers is a multiple of 5. Thus the product of (5k)(a) is divisible by 5 for any natural numbers a and k. Let us now answer the problem. The 2-element subsets we are looking for is in the form {5k, a} (order does not matter). Thus, the smallest value of k is 1 and the largest is k = 200. Thus we consider 200 cases. Case 1. Let k = 1, thus {5k, a} = {5, a}. There are 999 subsets of this form. (All one thousand elements except 5). Case 2. Let k = 2, thus {5k, a} = {10, a}. Again, there are 999 subsets of this form. However, the subset {10, 5} is already counted in the first case. Thus we only count the subsets of this form that are not in the previous case. There are 998 such subsets. (All one thousand elements except 5 and 10). .. . Case 200. Let k = 200 thus {5k, a} = {1000, a}. There are 800 subsets that are distinct from the previous cases. (All one thousand elements except 5, 10, 15, . . . , 990, 995 and 1000). Thus there are 999 + 998 + 997 + · · · + 801 + 800 = 179900 2-element subsets of X whose product of the elements is divisible by 5. TT We shall partition the set X into two sets: X1 = {5, 10, 15, . . . , 995, 1000} and X2 = {1, 2, 3, 4, 6, . . . , 997, 998, 999}. Note that all elements in X1 are divisible by 5 while those in X2 are not. Furthermore, |X1 | = 200 and |X2 | = 800. To get two element subsets of X such that the product its elements is divisible by 5, we either get two elements from set X1 , which can be done in 200 2 800 ways, or get one element each from set X1 and X2 , which can be done in 200 ways. Thus 1 1 800 by AP, the total number of such subsets is 200 + 200 = 179900. 1 1 2 33. Consider the following set of points in the xy-plane: A = {(a, b) | a, b ∈ Z and |a| + |b| ≤ 2}. Find Exercises 1
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(i) |A|; (ii) the number of straight lines which pass through at least 2 points in A; and (iii) the number of triangles whose vertices are points in A. 34. Let P be a convex n-gon, where n ≥ 6. Find the number of triangles formed by any 3 vertices of P that are pairwise nonadjacent in P . 35. 6 boys and 5 girls are to be seated around a table. Find the number of ways that this can be done in each of the following cases: (i) There are no restrictions; PB (11 − 1)! = 10! ways. (ii) No 2 girls are adjacent; PB We first arrange the 6 boys around the table, then insert the girls in between the boys. This guarantees that no two girls will be seated next to each other. Hence we have: 5! 65 5! ways. (iii) All girls form a single block; PB We first chunk all the girls and treat them as one “object”. We now have a total of 7 “objects”, which we have to arrange around the table. We then permute the girls inside the chunk. Hence we have: 6!5! ways. (iv) A particular girl G is adjacent to two particular boys B1 and B2 . PB We first chunk [B1 GB2 ], so now we have a total of 9 “objects”, which we have to arrange around the table. Then we permute B1 and B2 . Hence we have: 8!2! ways. 36. Show that the number of r-circular permutations of n distinct objects, where 1 ≤ r ≤ n, is given n! by (n−r)!·r . 37. Let k, n ∈ N. Show that the number of ways to seat kn people around k distinct tables such that there are n people in each table is given by (kn)! . nk 38. Let r ∈ N such that
1 1 9 − 10 = r
r
11 6
11 r
.
Find the value of r. FO 11 r!(9 − r)! r!(10 − r)! (11)(r!)(11 − r)! ⇐⇒ − = 10! 9! 10! 6(11!) 6( r!(11−n)! ) r!(9 − r)! (10 − r) (11)(r!)(11 − r)! ⇐⇒ 1− = 9! 10 6(11!) 9! r!(9 − r)! r (11)(r!)(11 − r)(10 − r)(9 − r)! 9!(10) ⇐⇒ = r!(9 − r)! 9! 10 6(11)(10)(9!) r!(9 − r)! (11 − r)(10 − r) ⇐⇒ r = ⇐⇒ 6r = 110 − 21r + r2 ⇐⇒ r2 − 27r + 110 = 0 6 ⇐⇒ (r − 22)(r − 5) = 0 ⇐⇒ r = 22 or r = 5 But it is impossible for nr where r > n — by definition, r ≤ n. 1
9! r!(9−r)!
−
1
10! r!(10−r)!
=
Thus, r = 5. 39. Prove each of the following identities: n n n−1 (a) = , where n ≥ r ≥ 1; r r r−1 LL n n−1 n (n − 1)! n! n = = = . r r−1 r (r − 1)!(n − 1 − r + 1)! r!(n − r)! r Exercises 1
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n n−r+1 n (b) = , where n ≥ r ≥ 1; r r r−1 LL n−r+1 n n−r+1 n! n!(n − r + 1) n! n = = = = . r r−1 r (r − 1)!(n − r + 1)! r!(n − r + 1)! r!(n − r)! r n n n−1 (c) = , where n > r ≥ 0; r n−r r LL n n−1 n (n − 1)! n! n = = = . n−r r n − r r!(n − 1 − r)! r!(n − r)! r n m n n−r = , where n ≥ m ≥ r ≥ 0. (d) m r r m−r LL n n−r n! (n − r)! n!(n − r)! m! = = · r m−r r!(n − r)! (m − r)!(n − r − m + r)! r!(n − r)!(m − r)!(n − m)! m! n! m! n m = · = . m!(n − m)! r!(m − r)! m r 40. Prove the identity
n r
=
n n−r
by using the bijection principle.
41. Let X = {1, 2, . . . , n}, A = {A ⊂ X | n ∈ / A}, and B = {A ⊂ X | n ∈ A}. Show that |A| = |B| by using the bijection principle. MS For every A ∈ A, X \ A ∈ B; for every B ∈ B, X \ B ∈ A. Therefore, |A| = |B|. 42. Let r, n ∈ N. Show that the product (n + 1)(n + 2) · · · (n + r) of r consecutive positive integers is divisible by r!. = r! n+r MS (n + 1)(n + 2) · · · (n + r) = Pn+r r r . Hence, r! | (n + 1)(n + 2) · · · (n + r). 43. Let A be a set of kn elements, where k, n ∈ N. A k-grouping of A is a partition of A into k-element subsets. Find the number of different k-groupings of A. KM For the first k-subset, kn 2nd k-subset, kn−k . For the 3rd k-subset, kn−2k . And k .For the k k k for the nth k-subset, kn−(n−1)k = . Hence, the number of different k-groupings would seem k kn−k kn−2k k k to be kn · · · k . But we still need to divide by the number of ways to arrange the k k k n k-subsets sincethe the ordering of the k-subsets doesn’t matter. Therefore, the final answer is kn kn−k kn−2k k · · · k k k k . n! 44. Twenty-five of King Arthur’s knights are seated at their customary round table. Three of them are chosen — all choices of three being equally likely — and are sent off to slay a troublesome dragon. Let P be the probability that at least two of the three had been sitting next to each other. If P is written as a fraction in lowest terms, what is the sum of the numerator and denominator? KM Number in sample space = 25 3 = 2300. How many ways can we choose 3 knights such that at least two of the three had been sitting next to each other (event E)? Case 1: There are exactly 2 knights sitting next to each other. There are 25 choices for the first knight, there are 2 ways to choose a seatmate (his left or right), and there are 21 ways to select another knight which is not adjacent to either of the knights already chosen. Divide by 2! since we counted twice the possibility that the seatmate was chosen first and the original was chosen next. Ans: (25)(2)(21) = 525. 2! Case 2: There are exaclty 3 knights sitting next to each other. Ans: There are 25 such consecutive ”triples”. Exercises 1
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Therefore, P (E) = 57.
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525 + 25 550 11 = = and the sum of the numerator and the denominator is 2300 2300 46
45. One commercially available ten-button lock may be opened by depressing — in any order — the correct five buttons. The sample shown below has {1, 2, 3, 6, 9} as its correct combination. Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations would this allow? 1 2 3 4 5
a a a a a a a a a
a a a
a a a
6 7 8 9 10
MS Currently, the locks require five buttons to unlock, which indicates that there are 10 5 combinations. If, after redesigning, as few as one button to as many as nine buttons can be used as a combination, only two combinations will not be viable: the combination where no button is 10 depressed 10 , and the combinations where all buttons are depressed . Hence, the number 0 1010 10 10 10 10 10 10 of combinations added by the redesign is 1 + 2 + 3 + 4 + 6 + 7 + 10 8 + 9 = 770. 46. In a shooting match, eight clay targets are arranged in two hanging columns of three each and one column of two, as pictured. A marksman is to break all eight targets accoring to the following rules: (1) The marksman first chooses a column from which a target is to be broken. (2) The marksman must then break the lowest remaining unbroken target in the chosen column. If these rules are followed, in how many different orders can the eight targets be broken?
i i i
i i
i i i
TT Consider a string of letters that will be composed of three As, two Bs and three Cs. The number 8! of ways in which we can construct such a string is equal to 3!·2!·3! . Now let’s note a particular string, say ABACCABC. If we are to relate this to the given problem, we can treat the three As as the clay targets on the first column, the two Bs as the clay targets on the second column, and the three Cs as the clay targets on the third column. Considering the given arrangement of letters in the string, this can be interpreted as: (1) on the first shot, the first (or bottom) clay target on the first column was hit; (2) on the second shot, the first (or bottom) clay target of the second column was hit; (3) on the third shot, the second (or middle) clay target of the first column was hit; (4) on the fourth shot, the first (or bottom) clay target of the third column was hit; (5) on the fifth shot, the second (or middle) clay target of the third column was hit; (6) on the sixth shot, the third (or top) clay target of the first column was hit; (7) on the seventh shot, the second (or top) clay target of the second column was hit; (8) and finally on the last shot, the remaining clay target on the third column was hit. Note that the interpretation is still consistent with the rules stated in the problem. Therefore, we can say that there is a bijection between the number of ways in which we can arrange three As, two Bs and three Cs to form a string, and the number of ways in which a shooter can break all the 8 clay targets according to the rules. Exercises 1
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Thus the number of ways a shooter can accomplish the task is equal to
8! 3!·2!·3!
= 560.
47. Using the numbers 1, 2, 3, 4, 5 we can form 5!(= 120) 5-digit numbers in which the 5 digits are all distinct. If these numbers are listed in increasing order: 12345, 12354, 12435, . . . , 54321, 1st
2nd
3rd
120th
find (i) the position of the number 35421; (ii) the 100th number in the list. FO
12345 12354 12435 12453 12534
12543 13245 13254 13425 13452
13524 13542 14235 14253 14325
14352 14523 14532 15234 15243
15324 15342 15423 15432 21345
21354 21435 21453 21534 21543
23145 23154 23415 23451 23514
23541 24135 24153 24315 24351 24513 24531 25134 25143 25314 25341 25413 25431 31245 31254 31425 31452
31524 31542 32145 32154 32415 34251 32514 32541 34125 34152 34215 34251 34512 34521 35124 35142 35214
35241 35412 35421 41235 41253 41325 41352 41523 41532 42135 42153 42315 42351 42513 42531 43125 43152
43215 43251 43512 43521 45123 45132 45213 45231 45312 45321 51234 51243 51324 51342 51423 52134 52143
52341 52314 52413 52431 53124 53142 53214 53241 54123 54132 54213 54231 54312 54321
(i) The position of the number 35421 is in the 72nd position. (ii) the 100th number in the list is the number 51342. 48. The P43 (= 24) 3-permutations of the set {1, 2, 3, 4} can be arranged in the following way, called the lexicographic ordering: 123, 124, 132, 134, 142, 143, 213, 214, 231, 234, 241, 243, 312, . . . , 431, 432. Thus the 3-permutations “132” and “214” appear at the 3rd and 8th positions of the ordering respectively. There are P94 (= 3024) 4-permutations of the set {1, 2, . . . , 9}. What are the positions of the 4-permutations “4567” and “5182” in the corresponding lexicographic ordering of the 4permutations of the set {1, 2, . . . , 9}? KM Let us consider first 4567. We will find all the permutations less than 4567. There are (3)(8)(7)(6) = 1008 numbers less than 4000. (The thousands place can only come from {1, 2, 3}.) There are (1)(3)(7)(6) = 126 numbers greater than 4000 but less than 4500. (The hundreds can only be {1, 2, 3}. The thousands place is already 4.) There are (1)(1)(3)(6) = 18 numbers greater than 4500 but less than 4560. (The tens can only be {1, 2, 3}.) Lastly, there are (1)(1)(1)(3) = 3 numbers greater than 4560 but less than 4567. (The ones can come from {1, 2, 3}.) Thus, the total number of permutations less than 4567 totals to 1155, implying that 4567 is the 1156th permutation in lexicographic ordering. Finding the position of 5182 is done similarly. There are (4)(8)(7)(6) = 1344 numbers less than 5000. (The thousands can be {1, 2, 3, 4}.) There are no permutations greater than 5000 but less than 5100. There are (1)(1)(5)(6) = 30 numbers greater than 5100 and less than 5180. (The tens can come from {2, 3, 4, 6, 7} and the ones can come from those remaining from the choices for the tens (4), along with 8 and 9. This gives 6 choices for the ones.) Lastly, there are no permutations greater than 5180 but less than 5182. Thus, the total number of permutations less than 5182 totals to 1374, implying that 5182 is the 1375th permutation in lexicographic ordering.
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49. The 53 (= 10) 3-element subsets of the set {1, 2, 3, 4, 5} can be arranged in the following way, called the lexicographic ordering: {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}. Thus the subset {1, 3, 5} appears in the 5th position of the ordering. There are 10 4 4-element subsets of the set {1, 2, . . . , 10}. What are the positions of the subsets {3, 4, 5, 6} and {3, 5, 7, 9} in the corresponding lexicographic ordering of the 4-element subsets of {1, 2, . . . , 10}? PB Notice that there are six 3-element subsets of the form {1, a, b}, where a and b are distinct elements of the set {2, 3, 4, 5}. Since the subset {2, 3, 4} appears right next to these six subsets, then it should appear in the 7th position. Keeping this in mind, it would now be easy for us to determine the position of a given subset. There are 93 4-element subsets of the form {1, a, b, c}, where a, b and c are distinct elements of the set {2, 3, . . . , 10}; and there are 83 4-element subsets of the form {2, d, e, f }, where d, e and f are distinct elements of the set {3, 4, . . . , 10}. Also, thesubset {3, 4, 5, 6} is the first subset starting with 3. Hence the subset {3, 4, 5, 6} appears in the 93 + 83 + 1 = 141 th position. More so, there are 62 4-element subsets of the form {3, 4, g, h}, where g and h are distinct ele ments of the set {5, 6, . . . , 10}; and 41 4-element subsets of the form {3, 5, 6, i}, where i is an element of the set {7, 8, 9, 10}. Since the subset {3, 5, 7, 9} is the second subset of the form {3, 5, 7, j}, where j is an element of the set {8, 9, 10}, then it should appear in the position numbered 9 8 6 4 + + + + 2 = 161. 3 3 2 1 50. Six scientists are working on a secret project, They wish to lock up the documents in a cabinet so that the cabinet can be opened when and only when three or more of the scientists are present. What is the smallest number of locks needed? What is the smallest number of keys each scientist must carry? 51. A 10-storey building is to be painted with some 4 different colors such that each storey is painted with one color. It is not necessary that all the 4 colors must be used. How many ways are there to paint the building if (i) there are no other restrictions? PB Each floor has four choices, hence there are 410 ways to paint the building. (ii) any 2 adjacent stories must be painted with different colors? PB Suppose we start painting with the ground floor, then initially we have four choices. But for the second floor, we only have three choices not including our previous choice of color. Again, for the third floor, we only have three choices, and so on. Thus there are 4 · 39 ways to paint the building. 52. Find the number of all multi-subset of M = {r1 · a1 , r2 · a2 , . . . , rn · an }. TT To find the number of all multi-subsets of M = {r1 · a1 , r2 · a2 , . . . , rn · an }, note that, for every object ai , there are (1 + ri ) ways of “choosing” it — that is, the multisubset of M may contain none of object ai , only one of object ai , two of object ai , up untill all of object ai (for a total (1+ri ) ways in which object ai can be a part of the multi-subset). Thus by (MP), the total number of multi-subsets of M is equal to (1 + r1 )(1 + r2 ) · · · (1 + rn ). 53. Let r, n ∈ N with r ≤ n. A permutation x1 x2 · · · x2n of the set {1, 2, . . . , 2n} is said to have the property P (r) if |xi − xi+1 | = r for at least one i in {1, 2, . . . , 2n − 1}. Show that, for each n and r, there are more permutations with property P (r) than without. 54. Prove by combinatorial argument that each of the following numbers us always an integer for each n ∈ N: TT The following expressions can be seen as the number of ways of arranging nk objects, such Exercises 1
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that k are of type 1, k are of type 2, . . . , k are of type n. And since these expressions are a result of counting the number of arrangements, then they must always be an integer. Note that the number of ways to arrange such nk objects is given by (kn)! . k! · | k!{z· · · k!} n times
(i)
(3n)! , 2n · 3n TT
(3n)! (3n)! (3n)! = = . n n n 2 ·3 (3!) 3! | · 3!{z· · · 3!} n times
This can viewed as the number of ways of arranging 3n objects, such that 3 are of type 1, 3 are of type 2, . . . , 3 are of type n. (6n)! (ii) n 2n 4n , 5 ·3 ·2 TT (6n)! (6n)! (6n)! . = = n 2n 4n n 5 ·3 ·2 (6!) 6! | · 6!{z· · · 6!} n times
This can viewed as the number of ways of arranging 6n objects, such that 6 are of type 1, 6 are of type 2, . . . , 6 are of type n. (iii)
(n2 )! , (n!)n TT
(n2 )! (n2 )! = . (n!)n n! | · n! {z· · · n!} n times
This can viewed as the number of ways of arranging n2 objects, such that n are of type 1, n are of type 2, . . . , n are of type n. Note that the total number of objects, given you have n kinds, each with n objects, is n · n = n2 . (n!)! (iv) . (n!)(n−1)! TT (n!)! [n · (n − 1)!]! . = (n−1)! n! (n!) | · n! {z· · · n!} (n−1)! times
This can viewed as the number of ways of arranging n! objects, such that n are of type 1, n are of type 2, . . . , n are of type (n − 1)!. Note that the total number of objects, given you have (n − 1)! kinds, each with n objects, is n · (n − 1)! = n!. 55. Find the number of r-element multi-subsets of the multi-set M = {1 · a1 , ∞ · a2 , ∞ · a3 , . . . , ∞ · an }. TT An r-element multi-subset of the multi-set M = {1 · a1 , ∞ · a2 , . . . , ∞ · an } may either (1) contain a1 , or (2) not contain a1 . Let xi represent the number of object ai in the subset. To determine the number of subsets in (1), we only need to consider the number of non-negative integer solutions to the equation r+n−3 x1 + x2 + · · · + xn = r, where x1 = 1. This is just equal to (r−1)+(n−1)−1 = r−1 r−1 . To determine the number of subsets in (2), we determine the number of non-negative integer solutions to the r+n−2 equation x2 + · · · + xn = r. This is just equal to (r)+(n−1)−1 = . r r Hence by (AP), the total number of r-element multi-subsets of M is equal to r+n−3 + r+n−2 . r−1 r Exercises 1
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56. Six distinct symbols are transmitted through a communication channel. A total of 18 blanks are to be inserted between the symbols with at least 2 blanks between every pair of symbols. In how many ways can the symbols and the blanks be arranged? PB We first permute the six distinct symbols then fix the ten blanks in such a way that exactly two blanks are between every pair of symbols. This leaves us with eight more blanks to be inserted in five spaces. Therefore the number of ways to arrange the symbols and the blanks is given by: 8+5−1 5−1 6! = 356400. 57. In how many ways can the following 11 letters: A, B, C, D, E, F, X, X, X, Y, Y be arranged in a row so that every Y lies between two Xs (not necessarily adjacent)? PB We first fix the three Xs. Since we want every Y to be in between two Xs, then we insert the Ys into the spaces between the Xs. Next, we insert the remaining six letters anywhere between the Xs and Ys including the “outside” spaces (space to the left of the leftmost X to the right of the and 6+6−1 rightmost X). Finally, we permute these six letters. Hence we have: 2+2−1 6!. 2 6 58. Two n-digit integers (leading zero allowed) are said to be equivalent if one is a permutation of the other. For instance, 10075, 01057 and 00751 are equivalent 5-digit integers. (i) Find the number of 5-digit integers such that no two are equivalent. TT The number of 5-digit integers such that no two are equivalent is just the same as the number of 5-element multi-subsets of M = {∞ · 0, ∞ · 1, . . . , ∞ · 9}, or the number of nonnegative integer solutions to the equation x0 + x1 +· · · + x9 = 5, where xi denotes the number of times digit i was used, which is just equal to 14 5 . (ii) If the digits 5, 7, 9 can appear at most once, how many nonequivalent 5-digit integers are there? TT If 5, 7, 9 can appear at most once, we shall consider four cases: (a) none of 5, 7, 9 are included The number of such 5-digit of non-negative integer solu P9 integers is equal to the number tions to the equation i=0 xi = 5. This will give us 11 numbers. 5 i6=5,7,9
(b) one of the numbers 5, 7, 9 is included The number of such 5-digit is equal to the number of non-negative integer solu P integers 9 tions to the equation i=0 xi + xs = 5, where xs = 1, for s = 5, 7 or 9. This will give i6=5,7,9 us 31 10 4 numbers. (c) two of the numbers 5, 7, 9 are included The number of such 5-digit is equal to the number of non-negative integer solu P numbers 9 tions to the equation i=0 xi + (xs + xt ) = 5, where xs = xt = 1, for s, t = 5, 7, 9 and i6=5,7,9 s > t. This will give us 32 93 numbers. (d) all of the numbers 5, 7, 9 are included The number of such 5-digit of non-negative integer solu P9 numbers is equal to the number tions to the equation i=0 xi = 2. This will give us 82 numbers. i6=5,7,9
Hence, the total number of such 5-digit numbers is equal to
11 5
+3
10 4
+3
9 3
+
8 2
.
59. How many 10-letter words are there using the letters a, b, c, d, e, f if (i) there are no restrictions? KM, LL 10-letter word, each letter has 6 choices. Ans: 610 . (ii) each vowel (a and e) appears 3 times and each consonant appears once? 10! . 3!3! LL The number of such 10-letter words is the number of permutations of the multi-set {3 · 10! a, b, c, d, 3 · e, f}, which is equal to 3!3! . KM Just count all possible arrangements of the word aaaeeebcdf. Ans:
Exercises 1
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(iii) the letters in the word appear in alphabetical order? KM This is bijective to looking forall nonnegative integer solutions to x1 + x2 + · · · + x6 = 10. 10 + 6 − 1 15 Ans: H610 = = . 10 10 (iv) each letter occurs at least once and the letters in the word appear in alphabetical order? KM This isbijective tolooking for all the positive integer solutions to x1 + x2 + · · · + x6 = 10. 4+6−1 9 6 Ans: H4 = = . 4 4 60. Let r, n, k ∈ N such that r ≥ nk. Find the number of ways of distributing r identical objects into n distinct boxes so that each box holds at least k objects. FO
k k k k ··· 1 2 3 n First put k objects in each box to fulfill the requirements. This can be done in one way. Then distribute the remaining r − nk objects in the boxes in an arbitrary way. Using (MP) and result in case 2(ii) r − nk + n − 1 r − n(k − 1) − 1 = ; r − nk r − nk by identity (1.4.2) this can also be written as r − n(k − 1) − 1 . n−1
61. Find the number of ways of arranging the 9 letters r, s, t, u, v, w, x, y, z in a row so that y always lies between x and z (x and y, or y and z need not be adjacent in the row). PB We first fix x, y and z respectively so that y is always between x and z. We then permute x and z. Next we insert the remaining six letters to the spaces between x, y and z, including the 6!. “outside” spaces. Finally we permute these six letters. Hence we have: 2! 4+6−1 6 62. Three girls A, B and C, and nine boys are to be lined up in a row. In how many ways can this be done if B must lie between A and C, and A, B must be separated by exactly 4 boys? TT Since B is in between A and C, then the girls can only be arranged in two ways: CBA and ABC. Now consider a particular arrangement, ABC. 9 boys must be placed in between/beside the girls; assume for now that every boy is the same. Since there should be exactly 4 boys in between A and B, we are going to have AY Y Y Y BC (Y represents a boy). Next we’re going to place the boys to the left of A, between B and C, and to the right of C. There are 3 spaces where the 5 remaining boys can position themselves in, and the number of ways in which they can to that is given by 5+3−1 . Since the 9 boys are different from 5 5+3−1 one another, by MP we shall have · 11!. 5 Since the number of ways of arranging the boys satisfying the given conditions given the initial arrangement of girls ABC is the same as that for CBA, then the total number of arrangements is just given by 5+3−1 · 11! · 2 = 42 · 11!. 5 63. Five girls and eleven boys are to be lined up in a row such that, from left to right, the girls are in the order G1 , G2 , G3 , G4 , G5 . In how many ways can this be done if G1 and G2 must be separated by at least 3 boys, and there is at most one boy between G4 and G5 ? 64. Given r, n ∈ N with r ≥ n, let L(r, n) denote the number of ways of distributing r distinct objects into n identical boxes so that no box is empty and the objects in each box are arranged in a row. Find L(r, n) in terms of r and n. r−1 . Multiply by KM The number of ways to place r identical objects into n distinct boxes is n−1 r! since the objects must be distinct and the arrangement matters. Divide by n! since the boxes must be identical and hence, its arrangements must be removed from the counting. Therefore, r! r−1 L(r, n) = n! n−1 . Exercises 1
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65. Find the number of integer solutions to the equation: x1 + x2 + x3 + x4 + x5 + x6 = 60 in each of the following cases: (i) xi ≥ i − 1 for each i = 1, 2, . . . , 6; PB By subtracting i − 1 to xi for each i = 1, 2, . . . , 6, we have adjusted the lower bound for each xi to zero instead P6 of i − 1. However, for the original equation to remain constant, we also have to subtract i=1 (i − 1) from 60. Hence the number of integer solutions to the equation given above is the same to the number of nonnegative integer solutions to the equation:
Hence we have
45+6−1 45
y1 + y2 + y3 + y4 + y5 + y6 = 45. = 50 5 .
(ii) x1 ≥ 2, x2 ≥ 5, 2 ≤ x3 ≤ 7, x4 ≥ 1, x5 ≥ 3 and x6 ≥ 2. PB Again we have to adjust each lower bound to zero instead of some other number for our general result to work. However, notice that x3 has an upper bound. Let us ignore this first for the meantime. By the same technique shown above, the number of integer solutions to the equation above (ignoring x3 ’s upper bound) is the same as the number of nonnegative integer solutions to the equation: y1 + y2 + y3 + y4 + y5 + y6 = 60 − 2 − 5 − 2 − 1 − 3 − 2 = 45. Hence we have 50 when x3 ≥ 8 and there 5 . However, we still have to remove allthe cases 44 50 44 are 39+6−1 = of them. Hence our final answer is − . 39 5 5 5 66. Find the number of integer solutions to the equation: x1 + x2 + x3 + x4 = 30 in each of the following cases: (i) xi ≥ 0 for each i = 1, 2, 3, 4; PB From our general result, the number of nonnegative integer solutions to the equation above is given by: 30+4−1 = 33 30 3 . (ii) 2 ≤ x1 ≤ 7 and xi ≥ 0 for each i = 2, 3, 4; PB By the same technique shown in the previous number, we have: 31 25 3 − 3 .
28+4−1 28
−
22+4−1 22
=
(iii) x1 ≥ −5, x2 ≥ −1, x3 ≥ 1 and x4 ≥ 2. PB Again, the number of integer solutions to this equation is the same as the number of nonnegative integer solutions to the equation: y1 + y2 + y3 + y4 = 30 + 5 + 1 − 1 − 2 = 33. Hence we have: 33+4−1 = 36 33 3 . 67. Find the number of quadruples (w, x, y, z) of nonnegative integers which satisfy the inequality w + x + y + z ≤ 1992. TT, JA The given inequality can be converted to an equation by introducing a slack variable s, where s is nonnegative: w + x + y + z + s = 1992. The slack variable s captures any excess value not taken up by the original variables. That means when w + x + y + z = 1991, then s = 1; when w + x + y + z = 1987, then s = 5. Since s is non-negative, it would not be possible for w + x + y + z > 1992, as this will force s to become negative. Thus the number of quadruplets (w, x, y, z) satisfying the inequality is the same the number of nonnegative integer solutions for the equation w + x + y + z + s = 1992. This can be interpreted as the number of ways of arranging 1992 sticks and 4 “+” signs, is equal to 1996 . 4 Exercises 1
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68. Find the number of nonnegative integer solutions to the equation: 5x1 + x2 + x3 + x4 = 14.
JA, LL Consider the following cases: Case 1: What if x1 = 0? Then the equation becomes x2 + x3 + x4 = 14. In this case, we have
n+r−1 r−1
=
14 + 2 2
=
16 . 2
Case 2: What if x1 = 1? Then the equation becomes 5(1) + x2 + x3 + x4 = 14, which is equal to x2 + x3 + x4 = 9, and by the same reasoning, we have n+r−1 9+2 11 = = . r−1 2 2 Case 3: What if x1 = 2? Then the equation becomes 5(2) + x2 + x3 + x4 = 14 ⇔ x2 + x3 + x4 = 4. In this case, we have
4+2 6 n+r−1 = = . 2 2 r−1
So, by A.P., we have,
16 11 6 + + = 120 + 55 + 15 = 190. 2 2 2
69. Find the number of nonnegative integer solutions to the equation: rx1 + x2 + · · · + xn = kr, where k, r, n ∈ N. JA, LL Consider the following cases: Case 1: What if x1 = 0? Then the equation becomes x2 + · · · + xn = kr. In this case, we have n − 2 pluses and so kr + n − 2 . n−2 Case 2: What if x1 = 1? Then the equation becomes r(1) + x2 + · · · + xn = kr which is equal to x2 + · · · + xn = kr − r = r(k − 1) and by the same reasoning, we have
Exercises 1
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We do this inductively until we reach this case: Case k + 1: What if x1 = k? Then the equation becomes r(k) + x2 + · · · + xn = kr which is equal to x2 + · · · + xn = kr − kr = r(k − k) and by the same reasoning, we have r(k − k) + n − 2 n−2 = . n−2 n−2 Getting a general form of this (after some thought) we have, k X r(k − i) + n − 2 n−2
i=0
.
70. Find the number of nonnegative integer solutions to the equation: 3x1 + 5x2 + x3 + x4 = 10.
PB Since each xi ≥ 0 for i = 1, 2, 3, 4, then 0 ≤ x1 ≤ 3 and 0 ≤ x2 ≤ 2. We now consider 4 · 3 = 12 cases: (x1 , x2 ) # of nonnegative integer solutions 10+2−1 11 (0, 0) = 10 1 = 11 7+2−1 8 (1, 0) = 7 1 = 8 4+2−1 5 (2, 0) = 4 1 = 5 1+2−1 2 (3, 0) = 1 1 = 2 5+2−1 6 (0, 1) 5 = 1 = 6 0+2−1 (0, 2) = 11 = 1 0
(x1 , x2 ) # of nonnegative integer solutions 2+2−1 3 (1, 1) = 2 1 =3 (1, 2) ∅→0 (2, 1) ∅→0 (2, 2) ∅→0 (3, 1) ∅→0 (3, 2) ∅→0
Hence there are 36 nonnegative integer solutions to the equation above. 71. Find the number of positive integer solutions to the equation: (x1 + x2 + x3 )(y1 + y2 + y3 + y4 ) = 77.
JA Consider the following cases: Case 1: If x1 + x2 + x3 = 7 and y1 + y2 + y3 + y4 = 11, then the number of nonnegative integer solutions to the equation x1 + x2 + x3 = 7 is given by r+n−1 7+3−1 9 9 = = = , r 7 7 2 and to the equation y1 + y2 + y3 + y4 = 11 is given by r+n−1 11 + 4 − 1 14 14 = . = = r 11 11 3 By M.P., the desired number is given by 9 14 . 2 3
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Case 2: If x1 + x2 + x3 = 11 and y1 + y2 + y3 + y4 = 7, then the number of nonnegative integer solutions to the equation y1 + y2 + y3 + y4 = 7 is given by r+n−1 7+4−1 10 10 = = = , r 7 7 3 and to the equation x1 + x2 + x3 = 11 is given by r+n−1 11 + 3 − 1 13 13 = = = . r 11 11 2 By M.P., the desired number is given by
10 3
13 . 2
By A.P., the desired number of nonnegative integer solutions to the equation (x1 + x2 + x3 )(y1 + y2 + y3 ) = 77 is given by
10 13 9 14 + . 3 2 2 3
72. Find the number of nonnegative integer solutions to the equation: (x1 + x2 + · · · + xn )(y1 + y2 + · · · + yn ) = p, where n ∈ N and p is a prime. PB Since p is a prime, then only either x1 + x2 + · · · + xn = p and y1 + y2 + · · · + yn = 1 or x1 + x2 + · · · + xn = 1 and y1 + y2 + · · · + yn = p. Hence we have: p+n−1 1+n−1 1+n−1 p+n−1 p+n−1 + = 2n . p 1 1 p p 73. There are 5 ways to express “4” as a sum of 2 nonnegative integers in which the order counts: 4 = 4 + 0 = 3 + 1 = 2 + 2 = 1 + 3 = 0 + 4. Given r, n ∈ N, what is the number of ways to express r as a sum of n nonnegative integers in which the order counts? 74. There are 6 ways to express “5” as a sum of 3 positive integers in which the order counts: 5 = 3 + 1 + 1 = 2 + 2 + 1 = 2 + 1 + 2 = 1 + 3 + 1 = 1 + 2 + 2 = 1 + 1 + 3. Given r, n ∈ N with r ≥ n, what is the number of ways to express r as a sum of n positive integers in which the order counts? 75. A positive integer d is said to be ascending if in its decimal representation: d = dm dm−1 · · · d2 d1 , we have 0 < dm ≤ dm−1 ≤ · · · ≤ d2 ≤ d1 . For instance, 1337 and 2455566799 are ascending integers. Find the number of ascending integers which are less than 109 . TT For this problem, we can only have ascending integers with at most nine digits. To determine all such integers, let us consider a seqeunce of 9 sticks and 9 “+” signs. If we were to initially place all nine “+” signs in a row, there will be 10 spaces available for the sticks to occupy (8 spaces in between the “+” signs and 2 spaces at the ends of the row). Relating this to the given problem, we can actually view the first 9 spaces as the numbers from 1 to 9 (i.e., the Exercises 1
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first space represents the digit 1, the 2nd space represents the digit 2, and so on). As for the 9 sticks, they determine the number of digits that the ascending number will have. In particular, if we let 7 of the sticks occupy (in any manner) the first 9 spaces — and therefore the 2 remaining sticks occupy the last space — this means that we’re actually forming a 7-digit number. If only 1 stick occupies any of the first 9 spaces — and therefore the 8 remaining sticks will occupy the last space — this means that we’re just forming a 1-digit number. Notice that the more sticks placed in the last space, the lesser the number of digits that the number will have. Moreover, we can only construct a number with at most 9 digits, since we only have nine sticks. To illustrate this, consider the sequence || + ||| + ||| + + + + + + + |. Using the above interpretation, we can say that the number contains 8 digits, composed of two 1s, three 2s, three 3s. Note, however, that there is only one way in which we can construct an ascending number, and that is 11222333. Here’s another sequence: + + | + +| + + + + + |||||||. This can be interpreted as an ascending number containing only two digits, and it is equal to 35. Thus, the number of ways of arranging 9 sticks and 9 “+” signs can give us the total number of ascending numbers less than 109 , which is 18 9 . However, we have to eliminate the sequence + + + + + + + + +||||||||||, since it does not correspond to any ascending number. Thus, the total number of ascending numbers less than 109 is
18 9
− 1.
76. A positive integer d is said to be strictly ascending if in its decimal representation: d = dm dm−1 · · · d2 d1 , we have 0 < dm < dm−1 < · · · < d2 < d1 . For instance, 145 and 23689 are strictly ascending integers. Find the number of strictly ascending integers which are less than (i) 109 , (ii) 105 . 77. Let A = {1, 2, . . . , n}, where n ∈ N. (i) Given k ∈ A, show that the number of subsets of A in which k is the maximum number is given by 2k−1 . (ii) Apply (i) to show that n−1 X
2i = 2n − 1.
i=0
78. In a given circle, n ≥ 2 arbitrary chords are drawn such that no three are concurrent within the interior of the circle. Suppose m is the number of points of intersection of the chords within the interior. Find, in terms of n and m, the number r of line segments obtained through dividing the chords by their points of intersection. 79. There are p ≥ 6 points given on the circumference of a circle, and every two of the points are joined by a chord. (i) Find the number of such chords. Assume that no 3 chords are concurrent within the interior of the circle. (ii) Find the number of points of intersection of these chords within the interior of the circle. (iii) Find the number of line segments obtained through dividing the chords by their points of intersection. Exercises 1
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(iv) Find the number of triangles whose vertices are the points of intersection of the chords within the interior of the circle. 80. In how many ways can n + 1 different prizes be awarded to n students in such a way that each student has at least one prize? FO Since there are n students, then they can be arranged n! ways for the prizes to be awarded. The fact that at least one prize each student then n+1 be number of distributions to be awarded r at n students, where r = 2, for r > 2 there will be student with no prize at all. Thus n+1 2 n! by (MP) are the total number of ways. 81.
(a) Let n, m, k ∈ N, and let Nk = {1, 2, . . . , k}. Find (i) the number of mappings from Nn to Nm . (ii) the number of 1-1 mappings from Nn to Nm , where n ≤ m. (b) A mapping f : Nn → Nm is strictly increasing if f (a) < f (b) whenever a < b in Nn . Find the number of strictly increasing mappings from Nn to Nm , where n ≤ m. (c) Express the number of mappings from Nn onto Nm in terms of S(n, m) (the Stirling number of the second kind).
82. Given r, n ∈ Z with 0 ≤ n ≤ r, the Stirling number s(r, n) of the first kind is defined as the number of ways to arrange r distinct objects around n identical circles such that each circle has at least one object. Show that (i) s(r, 1) = (r − 1)! for r ≥ 1; (ii) s(r, 2) = (r − 1)!(1 + 12 + 13 + · · · + (iii) s(r, r − 1) = 2r for r ≥ 2;
1 r−1 )
for r ≥ 2;
1 (iv) s(r, r − 2) = 24 r(r − 1)(r − 2)(3r − 1) for r ≥ 2; Pr (v) n=0 s(r, n) = r!.
83. The Stirling numbers of the first kind occur as the coefficients of xn in the expansion of x(x + 1)(x + 2) · · · (x + r − 1). For instance, when r = 3, x(x + 1)(x + 2) = 2x + 3x2 + x3 = s(3, 1)x + s(3, 2)x2 + s(3, 3)x3 ; and when r = 5, x(x + 1)(x + 2)(x + 3)(x + 4)
= 24x + 50x2 + 35x3 + 10x4 + x5 = s(5, 1)x + s(5, 2)x2 + s(5, 3)x3 + s(5, 4)x4 + s(5, 5)x5 .
Show that x(x + 1)(x + 2) · · · (x + r − 1) =
r X
s(r, n)xn ,
n=0
where r ∈ N. 84. Given r, n ∈ Z with 0 ≤ n ≤ r, the Stirling number S(r, n) of the second kind is defined as the number of ways of distributing r distinct objects into n identical boxes such that no box is empty. Show that (i) S(r, 2) = 2r−1 − 1; (ii) S(r, 3) = 12 (3r−1 + 1) − 2r−1 ; (iii) S(r, r − 1) = 2r ; (iv) S(r, r − 2) = 3r + 3 4r .
Exercises 1
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85. Let (x)0 = 1 and for n ∈ N, let (x)n = x(x − 1)(x − 2) · · · (x − n + 1). The Stirling numbers of the second kind occur as the coefficients of (x)n when xr is expressed in terms of (x)n s. For instance, when r = 2, 3 and 4, we have, respectively, x2 x3 x4
= = = = = =
x + x(x − 1) = (x)1 + (x)2 S(2, 1)(x)1 + S(2, 2)(x)2 x + 3x(x − 1) + x(x − 1)(x − 2) S(3, 1)(x)1 + S(3, 2)(x)2 + S(3, 3)(x)3 x + 7x(x − 1) + 6x(x − 1)(x − 2) + x(x − 1)(x − 2)(x − 3) S(4, 1)(x)1 + S(4, 2)(x)2 + S(4, 3)(x)3 + S(4, 4)(x)4
Show that for r = 0, 1, 2, . . . , xr =
r X
S(r, n)(x)n .
n=0
86. Suppose that m chords of a given circle are drawn in such a way that no three are concurrent in the interior of the circle. If n denotes the number of points of intersection of the chords within the circle, show that the number of regions divided by the chords in the circle is m + n + 1. 87. For n ≥ 4, let r(n) denote the number of interior regions of a convex n-gon divided by all its diagonals if no three diagonals are concurrent within the n-gon. as shown in the For instance, following diagrams, r(4) = 4 and r(5) = 11. Prove that r(n) = n4 + n−1 . 2
@
ZCCZZ C Z LlLl CC ,, L l ,l,C L, lC
@ @
88. Let n ∈ N. How many solutions are there in ordered positive integer pairs (x, y) to the equation xy = n? x+y PB Let n, x and y be elements of postive integers. Then the equation xy = n ⇐⇒ xy = n(x + y) ⇐⇒ xy − nx = ny ⇐⇒ x(y − n) = ny x+y
(88a)
Thus, ny > 0 and so as x(y − n) > 0. This implies that y − n > 0, which enables us to write y in terms of n, i.e., y = n + b, for some positive integer b. (88a) ⇐⇒ x =
ny , y−n
substituting n + a for y gives us: x=
n(n + b) n2 ⇐⇒ x = +n b b
(88b)
However, x is a positive integer, then for (88b) to be valid, then b must divide n2 , or in other words, b must be a factor of n2 . Thus this answers the problem: the number of solutions to the original problem is the same as the number of positive factors or divisors of n2 . Furthermore, let n = p1 k1 · p2 k2 · · · pm km , where pi s are distint prime numbers and ki s are positive integers. Then n2 would have (2k1 + 1)(2k2 + 1) · · · (2km + 1) factors. This is also the number of solutions to the problem given above. Exercises 1
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89. Let S = {1, 2, 3, . . . , 1992}. In each of the following cases, find the number of 3-element subsets {a, b, c} of S satisfying the given condition: (i) 3 | (a + b + c); PB Express S = {1, 2, 3, . . . , 1992} in terms of its remainder when divided by 3, that is, we can write it in its “modulo-3” form, say S 0 = {1, 2, 0, 1, 2, 0, . . . , 1, 2, 0}, where each number from S is represented in S 0 by its remainder when divided by 3. Thus we have 664 elements of 1s, 2s and 0s in S 0 . We denote the set of all 1s by set A, all 2s by set B, and all 0s by set C. Therefore, to have the sum of any 3 elements from S 0 divisible by 3, we can simply choose all three numbers from the one set, i.e., all from either set A, set B or set C; or we choose one number from each of the three sets, i.e., one number from set A, one from set B and one from set C. Therefore, the number of ways to do this is given by 3
3 664 664 . + 3 1
(ii) 4 | (a + b + c). PB Similarly, let S 00 = {1, 2, 3, 0, 1, 2, 3, 0, . . . , 1, 2, 3, 0}, where each number from S is represented in S 00 by its remainder when divided by 4. Thus we have 498 elements of 1s, 2s, 3s and 0s in S 00 . Again, we denote the set of all 1s by set A, all 2s by set B, all 3s by set C, and all 0s by set D. In order to have the sum of any three numbers from S 00 , we consider three cases. First, we can choose all three numbers from set D; second, we can choose exactly one number from set A, set C and set D, respectively; finally, we can choose two numbers from set C and one number from set B or two numbers from set A and one number from set B or two numbers from set B and one from set D. Therefore, the number of ways to do this is given by 3 498 498 498 498 + +3 . 3 1 1 2 90. A sequence of 15 random draws, one at a time with replacement is made from the set {A, B, C, . . . , X, Y, Z} of the English alphabet. What is the probability that the string UNIVERSITY occurs as a block in the sequence? PB The number of ways the string UNIVERSITY appears as a block is 265 · 6 and the number of ways to draw 15 letters with replacement is 2615 . Thus the probability is 265 · 6/2615 or simply 6/2610 . 91. A set S = {a1 , a2 , . . . , ar } of positive integers, where r ∈ N and a1 < a2 < · · · < ar , is said to be m-separated (m ∈ N) if ai − ai−1 ≥ m, for each i = 2, 3, . . . , r. Let X = {1, 2, . . . , n}. Find the number of r-element subsets of X which are m-separated, where 0 ≤ r ≤ n − (m − 1)(r − 1). MS It is evident that there is only one 0-element subset of X, ∅, and there are n 1-element subsets of X, and all these subsets are vacuously m-separated. Let x1 = a1 and xi = ai − ai−1 for 1 < i < r + 1 and xr+1 = n − ar . Note that x1 ≥ 1, xr+1 ≥ 0, and for 1 < i < r + 1, xi ≥ m and x1 + x2 + · · · + xr + xr+1 = n. There is a bijection between the collection of solutions S and the collection of solutions x1 , x2 , . . . , xr+1 ; there is also a bijection between the collection of solutions x1 , x2 , . . . , xr+1 and the collection of solutions y1 , y2 , . . . , yr+1 such that yi ≥ 0 for 1 ≤ i ≤ r + 1 and n−1−(r−1)m+r+1−1 y1 + · · · + yr+1 = n − 1 − (r − 1)m, which has cardinality Hr+1 = n−1−(r−1)m = n−1−(r−1)m n−(r−1)(m−1) n+m−1 n 1 2 . Note: when r = 0, Hn+m−1 = = 1; when r = 1, Hn−1 = 1 = n. r 0 92. Let a1 , a2 , . . . , an be positive real numbers, and let Sk be the sum of products of a1 , a2 , . . . , an taken k at a time. Show that 2 n Sk Sn−k ≥ a1 a2 · · · an , k Exercises 1
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for k = 1, 2, . . . , n − 1. TT First, we shall construct two sets S and T in the following manner: We will choose k positive integers from the set a1 , a2 , . . . , an , and get the product. Let that be s1 , which shall become an element in S. For the remaining n − k numbers not chosen, let their product be t1 . The resulting term shall become an element in T . Note that s1 · t1 = a1 a2 a3 · · · an . If we take a different set of k positive integers from a1 , a2 , . . . , an , we will again have to two products, one of which will become a member of S (say s2 ) and the other will become a member of T (say t2 ). Again, the product s2 · t2 = a1 a2 a3 · · · an . The process will continue until we have chosen all possible combinations of k positive integers from the set a1, a2 , . . . , an , which will allow us to fill in the sets S and T . Note that this process will be done nk times (referring to the number of possible ways k number may be chosen from n positive integers). Ultimately, we will have sets S = {s1 , s2 , . . . , s(n) } and T = {t1 , t2 , . . . , t(n) }, where si · ti = k k a1 a2 a3 · · · an for i = 1, 2, . . . , nk . Relating this to the given problem, we can actually say that Sn =
n (P k)
si . The addends that constitute
i=1
the sum Sn are always a product of k positive integers taken from a1 , a2 , . . . , an , which can be said for each element in set S. As for Sn−k , the products of n − k integers that make up its sum can be related to the elements of n set T (where each element is a product of n − k integers). While Sn−k is a sum of n−k terms, n we know that nk is equal to n−k . Thus, we can represent Sn−k as the sum of nk products, all n (P k) of which coming from the set T — that is, Sn−k = ti . i=1
To prove the inequality, we shall use two methods: the Generalized AM-GM Inequality and the Cauchy Schwarz Inequality. Method I: The Generalized AM-GM Inequality is stated as follows: For nonnegative numbers a1 , a2 , . . . , an , n
√ 1X ai ≥ n a1 a2 · · · an . n i=1 Since all elements of S and T are positive, then we can establish the following statements based on the Generalized AM-GM inequality, n
1/(k ) Sn ≥ s s · · · s n 1 2 n ( ) k
k
which can be converted to 1/(nk) n Sn ≥ s1 s2 · · · s(n) k k
n
1/(k ) Sn−k ≥ t t · · · t , n 1 2 n ( ) k
k
Sn−k
1/(nk) n ≥ t1 t2 · · · t(n) . k k
Combining the two inequalites, we shall have 2 1/(nk) n2 1/(nk) n n Sn Sn−k ≥ (s1 · t1 )(s2 · t2 ) · · · (s(n) · t(n) ) = (a1 a2 a3 · · · an )(k ) , k k k k since si · ti = a1 a2 a3 · · · an for i = 1, 2, . . . , nk ; then Sn Sn−k
2 n ≥ a1 a2 a3 · · · an , k
which is the desired inequality. Exercises 1
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Method II: The Cauchy-Schwarz Inequality is stated as follows: For real numbers x1 , x2 , . . . , xn , y1 , y2 , . . . , yn , we have n X
!2 xi yi
n X
≤
i=1
Since si , ti ≥ 0, then CS Inequality:
√
si and
√
! xi
2
i=1
n X
! yi
2
.
i=1
ti are real numbers for i = 1, 2, . . . ,
n k
. Hence, we can use
2 n n (k ) (k ) (nk) X √ √ X X si ti . si ti ≤
i=1
i=1
Note that si · ti = a1 a2 a3 · · · an for i = 1, 2, . . . , n 2 (k ) X √ si ti
n k
.
n n (k ) (k ) X X ti ≤ si
i=1
i=1
2
i=1
(nk) X √ a1 a2 a3 · · · an i=1
i=1
2 n √ a1 a2 a3 · · · an k 2 n a1 a2 a3 · · · an k
i=1
n n (k ) (k ) X X ≤ si ti i=1
n n (k ) (k ) X X ti si ≤ i=1
i=1
≤ Sn Sn−k ,
which is the desired inequality. 93. For {1, 2, 3, . . . , n} and each of its nonempty subsets, a unique alternating sum is defined as follows: Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. (For example, the alternating sum for {1, 2, 4, 6, 9} is 9 − 6 + 4 − 2 + 1 = 6 and for {5} it is simply 5.) Find the sum of all such alternating sums for n = 7. MS Let n ∈ N, A ⊆ Nn = {1, 2, . . . , n}, aA be the alternating sum of A, a∅ = 0, P(X) denote the power set of X, Nn = {A ∪ {n} : A ∈ P(Nn−1 )} = P(Nn ) \ P(Nn−1 ) and X An = aA — claim that An = n2n−1 . A∈P(Nn )
For each A ∈ P(Nn−1 ), there is a unique B = A ∪ {n} ∈ Nn ; for each B ∈ Nn , there is a unique A = B \ {n} ∈ P(Nn−1 ). If A ∈ P(Nn−1 ), then aA∪{n} = n − aA , since n is larger than the elements of A, and those elements will change signs relative to their sign in aA . Thus, since |P(Nk )| = 2k , X X X X X An = aA = aA + aB = (aA + aA∪{n} ) = n = n2n−1 . A∈P(Nn )
A∈P(Nn−1 )
B∈Nn
A∈P(Nn−1 )
A∈P(Nn−1 )
Therefore, the sum An of all such alternating sums for n = 7 is A7 = 7 · 27−1 = 448. 94. A gardener plants three maple trees, four oak trees and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let m n in lowest terms be the probability that no two birch trees are next to one another. Find m + n. 95. In a tournament, each player played exactly one game against each of the other players. In the game the winner was awarded 1 point, the loser got 0 points, and each of the two players earned 1/2 point if the game was a tie. After the completion of the tournament, it was found that exactly Exercises 1
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half of the points earned by each player were earned in games against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten.) What was the total number of players in the tournament? MS Let there be be n players in the tournament. Of the lowest ten players, 10 = 45 matches 2 were played, and thus, the lowest ten players scored a total of 45 points amongst themselves and 45 points with the other n − 10 players. There were n−10 matches played amongst the other 2 players, for as many total points, and n−10 + 45 consititutes half the total points of the n people in 2 the tournament, since the other half of the points were contributed by matches between people in the lowest ten and the others. There are n2 total matches, and thus the same number of points. Thus, n − 10 n n(n − 1) = 2 + 45 = (n − 10)(n − 11) + 90 = 2 2 2 n2 − n = 2n2 − 42n + 400 n2 − 41n + 400 = 0 =⇒ n = 16 or n = 25. If n = 16, then the top six players would only have an average of 2 62 /6 = 5 points, which would be less than the average score of the lowest ten players, which is 2 · 45/10 = 9 points. Thus, there were 25 players in the tournament. 96. Let S be the sum of the base 10 logarithms of all of the proper divisors of 1,000,000. (By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself.) What is the integer nearest to S? TT The prime factorization of 1,000,000 is given by 26 ·56 ; based on this, we can see that 1,000,000 has a total of 7 · 7 = 49 positive divisors. Let the divisors be denoted by (in no particular order of magnitude) d1 , d2 , . . . , d49 . Since S is the sum of the base 10 logarithms of all the proper divisors of 1,000,000 (which excludes 1 and 1,000,000), then S can be expressed as S = log d1 + log d2 + · · · + log d49 − log 1 − log 1000000 = log
d1 d2 · · · d49 . 1000000
To determine S, we should find the product d1 d2 · · · d49 . Note that each of the 49 divisors of 1,000,000 can be expressed in the form 2x · 5y , for 0 ≤ x, y ≤ 6. This means that the product d1 d2 · · · d49 can then be expressed in the form 2m ·5n , where m, n ∈ N. To solve for m, we consider that: (a) 7 of the divisors have the form 21 · 5y (21 · 50 , 21 · 51 , . . . , 21 · 56 ); (b) 7 of the divisors have the form 22 · 5y ; (c) 7 of the divisors have the form 23 · 5y ; (d) 7 of the divisors have the form 24 · 5y ; (e) 7 of the divisors have the form 25 · 5y ; and (f) 7 of the divisors have the form 26 · 5y . Considering all these cases, we can say that 2m = (21 )7 · (22 )7 · (23 )7 · (24 )7 · (25 )7 · (26 )7 = 2147 , which means m = 147. To solve for n, we may employ the same method, and by symmetry, we can show that 5n = (51 )7 · (52 )7 · (53 )7 · (54 )7 · (55 )7 · (56 )7 = 5147 , which means n = 147. Thus, the product of the divisors d1 d2 · · · d49 is equal to 2147 · 5147 = 10147 . Solving for the value of S, we have S = log
d1 d2 · · · d49 10147 = log = log 10141 = 141. 1000000 106
97. In a sequence of coin tosses one can keep a record of the number of instances when a tail is immediately followed by a head, a head is immediately followed by a head, etc. We denote these by T H, HH, etc. For example, in the sequence HHT T HHHHT HHT T T T of 15 coin tosses we observe that there are five HH, three HT , two T H and four T T subsequences. How many different sequencesof 15 coin tosses will contain exactly two HH, three HT , four T H and five T T subsequences? Exercises 1
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Mathematics 248
Topics in Combinatorial Mathematics
1st Semester AY 2006-2007
98. An ordered pair (m, n) of nonnegative integers is called simple if the addition m + n in base 10 requires no carrying, Find the number of simple ordered pairs of nonnegative integers that sum to (i) 1492; PB Let abcd and ef gh be two “simple” 4-digit numbers (the first digit can be zero, in that case, the number becomes 3-digit only). We want to find the number of “simple” integers that sum up to 1492. To do this, let abcd + ef gh = 1492. Since abcd and ef gh are “simple” integers, then no carrying is allowed in addition. Hence, the problem reduces to finding the number of non-negative integers to the following linear equations: a+e b+f c+g d+h
= 1 = 4 = 9 = 2
(98a) (98b) (98c) (98d)
It is easy to show that (98a) has 2 nonnegative integer solutions, (98b) has 5, (98c) has 10 and (98d) has 3. Thus, by multiplication property, there are a total of 2 · 5 · 10 · 3 = 300 number of simple nonnegative integer solutions to abcd + ef gh = 1492. (ii) 1992. PB Similarly, the problem reduces to finding the number of non-negative integers to the following linear equations: a+e b+f c+g d+h
= = = =
1 4 9 2
(98a) (98b) (98c) (98d)
(98a) has 2 non-negative integer solutions, (98b) has 10, (98c) has 10 and (98d) has 3. Thus, by multiplication property, there are a total of 2 · 10 · 10 · 3 = 600 number of simple nonnegative integer solutions to abcd + ef gh = 1992. 99. Let m/n, in lowest terms, be the probability that a randomly chosen positive integer of 1099 is an integer multiple of 1088 . Find m + n. PB 1099 can be expressed as 299 · 599 , thus 1099 has 1002 positive divisors. Integer multiples of 1088 can be expressed in the form 2x · 5y , where x and y are both integers and 88 ≤ x, y ≤ 99. Thus among the 10000 positive divisors of 1099 , 12 · 12 are integer multiples of 1088 . Therefore, the probability that a randomly chosen positive divisor of 1099 is an integer multiple of 1088 is 144/10000 or 9/625. Thus, m + n = 634. 100. A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron, one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face? MS It is evident that no two polygons of the same shape share a side, and thus there are 48 vertices on the polygon. Each vertex shares a face with 3 + 5 + 7 − 3 = 12 other vertices, since each of the faces that the vertex is on must share a side with an adjacent face. Thus, there are 48 − 13 = 35 other vertices that do not share a face with that vertex, and the edge connecting any of these vertices to the given point must lie in the interior of the polyhedron. Since each edge is counted twice (once per vertex it is incident on), the number of required edges must be 48 · 35/2 = 840. 101. Someone observed that 6! = 8 · 9 · 10. Find the largest positive integer n for which n! can be expressed as the product of n − 3 consecutive positive integers. MS It is apparent that the largest number in the n − 3 consecutive positive integers multiplied is
Exercises 1
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Mathematics 248
Topics in Combinatorial Mathematics
1st Semester AY 2006-2007
larger than n, and larger than 3. Let n + r be the largest number among the positive integers. Then the problem becomes solving for n in (n + r)(n + r − 1) · · · (n + r − (n − 4)) =
r r Y (n + r)! n! Y n + r n+r = n! =⇒ = n! =⇒ = 6. (r + 3)! 3! i=1 3 + r 3+r i=1
Since n > 3, the fraction (n + r)/(3 + r) > 1, and the larger r becomes, the smaller (n + 1)/4 becomes, and the smaller n becomes. Thus, n is largest when r = 1, and n + 1 = 24, n = 23. 102. Let S = {1, 2, . . . , n}. Find the number of subsets A of S satisfying the following conditions: A = {a, a + d, . . . , a + kd} for some positive integers a, d and k,
and
A ∪ {x} is no longer an arithmetic progression with common difference d for each x ∈ S \ A. 103. Find all natural numbers n > 1 and m > 1 such that 1! · 3! · 5! · · · (2n − 1)! = m!. 104. Show that for n ∈ N,
n X
Pnr = bn!ec,
r=0
where bxc denotes the greatest integer ≤ x and e = 2.718 . . . . PB First verify for n = 1: 1 X
Pnr =
r=0
1! 1! + = 2 = b1!(e)c. (1 − 0)! (1 − 1)!
For n ≥ 2, consider the power (Maclaurin) series of ex , ex = 1 + x +
x2 x3 xk xn xn+1 ξx + + ··· + + ··· + + e , where 0 < ξx < x. 2! 3! k! n! (n + 1)! {z } | error function
Letting x = 1 and multiplying everything by n!, n!e = n! +
n!
+
n! 2!
+···+
n! k!
= Pnn + Pnn−1 + Pnn−2 + · · · + Pnn−k
n! n! n! + + eξ1 (n − 1)! n! (n + 1)! n! +···+ Pn1 + Pn0 + eξ1 . (n + 1)!
+···+
However, n! eξ1 (n + 1)!
1 ξ1 1 e < e n+1 n+1 < 1
=
since
0 < ξ1 < 1
since
e < n + 1 for all n ≥ 2.
Thus, bn!ec =
Pnn
+
Pnn−1
+ ··· +
Pn1
+
Pn0
n X n! ξ1 + e = Pnn + Pnn−1 + · · · + Pn1 + Pn0 = Pnr . (n + 1)! r=0
105. Let S = {1, 2, . . . , 1990}. A 31-element subset A of S is said to be good if the sum by 5. Find the number of 31-element subsets of S which are good.
P
a∈A
a is divisible
106. Let S be a 1990-element set and let P be a set of 100-ary sequences (a1 , a2 , . . . , a100 ), where ai s are distinct elements of S. An ordered pair (x, y) of elements of S is said to appear in (a1 , a2 , . . . a100 ) if x = ai and x = aj for some i, j with 1 ≤ i < j ≤ 100. Assume that every ordered pair (x, y) of elements of S appears in at most one member in P. Show that |P| ≤ 800. Exercises 1
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Mathematics 248
Topics in Combinatorial Mathematics
1st Semester AY 2006-2007
107. Let M = {r1 · a1 , r2 · a2 , . . . , rm · an } be a multiset with r1 + r2 + · · · + rn = r. Show that the number of r-permutations of M is equal to the number of (r − 1)-permutations of M . r! TT The number of r-permutations of M is given by . As for the number of r − 1r1 !r2 ! · · · rm ! permutations of M , we shall consider the r − 1-permutations of several multisets of the form Mi = {r1 · a1 , r2 · a2 , . . . , (ri − 1) · ai , . . . , rm · am } for i = 1, 2, . . . , m, where |Mi | = r − 1. Counting the total number of permutations for all Mi , m X i=1
(r − 1)! = S. r1 !r2 ! · · · (ri − 1)! · · · rm !
Manipulating the expression, S
= = =
Since
m P
(r − 1)! (r − 1)! (r − 1)! + + ··· + (r1 − 1)!r2 ! · · · rm ! r1 !(r2 − 1)! · · · rm ! r1 !r2 ! · · · (rm − 1)! (r − 1)! 1 1 1 + + ··· + (r1 − 1)!(r2 − 1)! · · · (rm − 1)! r2 r3 · · · rm r1 r3 · · · rm r1 r2 · · · rm−1 r1 + r2 + · · · + rm (r − 1)! . (r1 − 1)!(r2 − 1)! · · · (rm − 1)! r1 r2 r 3 · · · rm
ri = r,
i=1
(r − 1)! S= (r1 − 1)!(r2 − 1)! · · · (rm − 1)!
r r1 r2 r 3 · · · rm
=
r! . r1 !r2 ! . . . rm !
Thus the number of r-permutations of M is equal to the number of r − 1-permutations of M . 108. Prove that it is impossible for seven distinct straight lines to be situated in the Euclidean plane so as to have at least six points where exactly three of these lines intersect and at least four points where exactly two of these lines intersect. 109. For what n ∈ N does there exist a permutation (x1 , x2 , . . . , xn ) of (1, 2, . . . , n) such that the differences |xk − k|, 1 ≤ k ≤ n, are all distinct? 110. Numbers d(n, m), where n, m are integers and 0 ≤ m ≤ n, are defined by d(n, 0) = d(n, n) = 1 for all n ≥ 0 and m · d(n, m) = m · d(n − 1, m) + (2n − m) · d(n − 1, m − 1) for 0 < m < n. Prove that all the d(n, m) are integers. 111. A difficult mathematical competition consisted of a Part I and a Part II with a combined total of 28 problems. Each contestant solved 7 problems altogether. For each pair of problems, there were exactly two contestants who solved both of them. Prove that there was a contestant who, in Part I, solved either no problems or at least four problems. 112. Suppose that five points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or coincident. From each point perpendiculars are drawn to all the lines joining the other four points. Determine the maximum number of intersections that these perpendiculars can have. 3
113. Let n distinct points in the plane be given. Prove that fewer than 2n 2 pairs of them are at unit distance apart. 114. If c and m are positive integers each greater than 1, find the number n(c, m) of ordered c-tuples (n1 , n2 , . . . , nc ) with entries from the initial segment {1, 2, . . . , m} of the positive integers such that n2 < n1 and n2 ≤ n3 ≤ · · · ≤ nc . Exercises 1
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Mathematics 248
Topics in Combinatorial Mathematics
1st Semester AY 2006-2007
115. Let X = {x1 , x2 , . . . , xm }, Y = {y1 , y2 , . . . , yn }, (m, n ∈ N) and A ⊆ X × Y . For xi ∈ X, let A(xi , ·) = ({xi } × Y ) ∩ A and for yi ∈ Y , let A(·, yi ) = (X × {yi }) ∩ A. (i) Prove the following Fubini Principle: m X
|A(xi , ·)| = |A| =
i=1
n X
|A(·, yj )|.
j=1
(ii) Using (i), or otherwise, solve the following problem: There are n ≥ 3 given points in the plane such that any three of them form a right-angled triangle. Find the largest possible value of n.
Source: Chen, Chuan-Chong and Koh, Khee-Meng. Principles and Techniques in Combinatorics. Reproduced without explicit permission, for use in Math 248 class, Ateneo de Manila University, Philippines, 1st Semester AY 2006-2007.
Exercises 1
74 of 115 solved
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